Question

A conducting rod of length l, rotates about one of its ends in a uniform magnetic field B, with a constant angular velocity ω. If the plane of rotation is perpendicular to B, the e.m.f. induced between the ends of rod is ______.

Options

• `1/2 B omega l^2`

• Bωl²

• 2Bωl²

• Bωl

Answer 1

A conducting rod of length l, rotates about one of its ends in a uniform magnetic field B, with a constant angular velocity ω. If the plane of rotation is perpendicular to B, the e.m.f. induced between the ends of rod is `bbunderline(1/2 B omega l^2)`.

Explanation:

Given: 

Length of the rod = l

Angular velocity = ω

Magnetic field = B

The plane of rotation is perpendicular to B.

Consider a small element of the rod at a distance r from the axis of rotation.

The linear velocity of this element is:

v = ωr

The small induced emf in this element is given by:

dE = Bv dr

= B(ωr) dr

To find the total induced emf between the ends of the rod, integrate from r = 0 to r = l:

E = `int_0^l Bomega r  dr`

= `B omega int_0^l r  dr`

= `B omega [r^2/2]_0^l`

= `1/2 B omega l^2`